Question:
A ray of light is projected from the origin at angle of \(60^{\circ}\) with the positive direction of \(x\)-axis towards the line, \(y=2\), which gets reflected from the point \((\alpha, 2)\). Then the distance of the reflected ray of light from the point \((2,2)\) is :
A ray of light is projected from the origin at angle of \(60^{\circ}\) with the positive direction of \(x\)-axis towards the line, \(y=2\), which gets reflected from the point \((\alpha, 2)\). Then the distance of the reflected ray of light from the point \((2,2)\) is :
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The reflected ray makes the supplementary angle with the x-axis; find its equation through the point of incidence and use the point-to-line distance formula.
Updated On: Nov 12, 2026
- \(3-\sqrt{3}\)
- \(\sqrt{3}-1\)
- \(1-\frac{1}{\sqrt{3}}\)
- \(2\left(1-\frac{1}{\sqrt{3}}\right)\)
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The Correct Option is B
Solution and Explanation
The incident ray from the origin has slope \(\tan 60^{\circ}=\sqrt{3}\), so its equation is \(y=\sqrt{3}x\).
It meets \(y=2\) at \(x=\frac{2}{\sqrt{3}}\), so \(\alpha=\frac{2}{\sqrt{3}}\).
On reflection from the horizontal line \(y=2\), the slope changes sign, so the reflected ray has slope \(-\sqrt{3}\) and passes through \(\left(\frac{2}{\sqrt{3}},2\right)\).
Equation: \(y-2=-\sqrt{3}\left(x-\frac{2}{\sqrt{3}}\right)\Rightarrow \sqrt{3}x+y-4=0\).
Distance of \((2,2)\): \(\frac{|2\sqrt{3}+2-4|}{\sqrt{3+1}}=\frac{2\sqrt{3}-2}{2}=\sqrt{3}-1\).
Answer: (2) \(\sqrt{3}-1\)
It meets \(y=2\) at \(x=\frac{2}{\sqrt{3}}\), so \(\alpha=\frac{2}{\sqrt{3}}\).
On reflection from the horizontal line \(y=2\), the slope changes sign, so the reflected ray has slope \(-\sqrt{3}\) and passes through \(\left(\frac{2}{\sqrt{3}},2\right)\).
Equation: \(y-2=-\sqrt{3}\left(x-\frac{2}{\sqrt{3}}\right)\Rightarrow \sqrt{3}x+y-4=0\).
Distance of \((2,2)\): \(\frac{|2\sqrt{3}+2-4|}{\sqrt{3+1}}=\frac{2\sqrt{3}-2}{2}=\sqrt{3}-1\).
Answer: (2) \(\sqrt{3}-1\)
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